ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A.P. HILLMAN University of Santa Clara, Santa Clara, California

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1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A.P. HILLMAN Uiversity of Sata Clara, Sata Clara, Califoria ' Sed all commuicatios regardig E l e m e t a r y P r o b l e m s ad Solutios to Professor A. P. Hillma, Mathematics Departmet, Uiv e r s i t y of Sata Clara, Sata Clara, Califoria. Ay problem believed to be ew i the a r e a of r e c u r r e t sequeces ad ay ew approaches to existig problems will be welcomed. The proposer should submit each problem with solutio i legible form, preferably typed i double spacig with ame ad a d d r e s s of the proposer as a headig. Solutios to p r o b l e m s listed below should be submitted o s e - p a r a t e siged sheets withi two moths of publicatio. B Proposed by Douglas Lid, Falls Church, Virgiia Prove that for every positive iteger k there a r e o m o r e tha k k+1 Fiboacci u m b e r s betwee ad B Proposed by Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas Let H be the -th geeralized Fiboacci umber, i. e., let H, ad H~ be a r b i t r a r y J ad H, = H,, + H for > 0. Show that H x + ( - l ) H 2 + (-2)H H = H ( + 2 ) H 2 - H 1. B Proposed by C.A. Church, Jr., Duke Uiversity, Durham, North Carolia Evaluate the -th order determiat a+b ab a+b ab a+b ab D = a+b 231

2 232 ELEMENTARY PROBLEMS AND SOLUTIONS October B Proposed by Barry Litvack, Uiversity of Michiga, A Arbor, Michiga Prove that for every positive iteger k there a r e k cosecutive Fiboacci umbers each of which is composite. ] 3-4 g Proposed by H.H. Fers, Uiversity, of Victoria, Victoria, British Columbia, Caada Prove that r 1 I r I -2 F 2 <- 2 >\> F k = k=i if r is a eve positive iteger r ( r - 1 ) / 2 2 F -2(5) ' if r is a odd positive iteger, where F 10 = F., + F (F. = F» = 1) ad fid the correspodig & sum * i which the F, a r e replaced by the Lucas u m b e r s L,. B Proposed by Ato Qlaser, Pesylvaia State Uiversity, Abigto, Pesylvaia Let <p r e p r e s e t the Letter "oh". Give that T, W,«, L, V, P, ad TW< a r e Fivoacci u m b e r s, solve the cryptarithm i the base 14, itroducig the digits a> /3 > d > ad S i base 14 for 10, 11, 12, ad 13 i base 10. TW0 IS THE 0NLY EVEN PRIME B Proposed by Douglas Lid, Falls Church, Virgiia Prove that j=0 2 F 2 - ( ) F. J F 2. B Proposed by Douglas Lid, Falls Church, Virgiia Let <f> () be the Euler totiet ad let <f> () be defied by <f> l () = 0 (), <f> k + 1 ( ) = <j> [^k()]. Prove that 0 ( F ) = 1, where F is the -th Fiboacci umber.

3 1964 E L E M E N T A R Y P R O B L E M S A N D SOLUTIONS 233 SOLUTIONS A PERIODIC R E C U R R E N T S E Q U E N C E B Proposed by V.E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Fid the millioth t e r m of the sequece a give that a, = 1, a-, = 1, ad a,~ = a., - a for > Solutio by J.A.H. Huter, Toroto, Otario, Caada It is simple to show that a has a period of 6, with: a 6k+4 = a 6k+5 = " 1 ' 10 = 4(mod 6), hece the millioth t e r m must be - 1. Also solved by Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas; Joh H. Halto, Uiversity of Colorado, Boulder, Colorado; J.L. Brow, Jr., Pesylvaia State Uiversity, State College, Pesylvaia; Vassili Daiev, Sea Cliff, L.I., N.Y.; George Ledi, Jr., Sa Fracisco, Califoria; Roald Weishek, Sa Jose State College, Sa Jose, Califoria; Dermott A. Breault, Sylvaia A.R.L., Waltham, Mass.; David E. Zitarelli, Temple Uiversity, Philadelphia, Peyslvaia; B. Litvack, Uiversity of Michiga, A Arbor, Michiga; ad the proposer. SUMS OF CONSECUTIVE FIBONACCI NUMBERS B Proposed by Douglas Lid, Falls Church, Virgiia If is eve, show that the sui of Z cosecutive Fiboacci u m b e r s is divisible by F. J Solutio by Roseaa Torretto, Uiversity of Sata Clara, Sata Clara, Califoria Let T be the sum F,, F,., of 2 cosecutive a+1 a+2 Fiboacci u m b e r s. Let S = F, + F~ F. It is well kow 1 2 that S = F,~ - 1. Hece +2 e a+2 ~ a+2+2 a+2 Sice F, - F = L F for p eve (see I. D. Ruggles, Some q+p q- p q P F && Fiboacci Results usig Fiboacci-Type Sequeces, this Quarterly, Vol. 1, No. 2, r * p. 77), ' T = L a++2,, - F as desired.

4 234 E L E M E N T A R Y P R O B L E M S A N D S O L U T I O N S O c t o b e r Also solved by J.L. Boriu, Jr., Pesylvaia State Uiversity, State College, Pesylvaia; B. hitvack, Uiversity of Michiga, A Arbor, Michiga; Joh H. Halto, Uiversity of Colorado, Boulder, Colorado; Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas; ad the proposer. A C O N G R U E N C E R E L A T I O N B Proposed by Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas Show t h a t L = F ( m o d 5). x ' Solutio by Joh Alle Fuchs, Uiversity of Sata Clara, Califoria It f o l l o w s f r o m b a s i c r e s u l t s o h o m o g e e o u s l i e a r d i f f e r e c e e q u a t i o s t h a t the s e q u e c e Y = L - F s a t i s f i e s ^. ^ (1) Y, A = 2Y, - + Y, 9-2Y,, - Y, x ' i. e., (E - E - 1) Y = 0 w i t h the o p e r a t o r E defied a s i J a m e s A., J e s k e, L i e a r R e c u r r e c e R e l a t i o s P a r t I, t h i s Q u a r t e r l y, Vol. 1, N o. 2. The d e s i r e d r e s u l t ow follows by t r i a l for = 1, 2, 3, a d 4 a d m a t h e m a t i c a l i d u c t i o u s i g (1). Also solved by Joh H. Halto, Uiversity of Colorado, Boulder' Colorado; J.L. Brow, Jr., Pesylvaia Slate Uiversity, State College, Pesylvaia; Douglas Lid, Falls Church, Virgiia; ad the proposer. T E R M B Y T E R M S U M S B Proposed by Joh A. Fuchs, Uiversity of Sata Clara, Sata Clara, Califoria L e t u, v,.., w be s e q u e c e s e a c h s a t i s f y i g the s e c o d ^ J O. o r d e r r e c u r r e c e f o r m u l a + y, r» = g y i y ( - i ) > J & + 2 / + l J N ' w h e r e g a d h a r e c o s t a t s. L e t a, b,..., c be c o s t a t s. Show t h a t a u + bv cw = 0 i s t r u e for a l l p o s i t i v e i t e g r a l v a l u e s of if it is t r u e for = 1 a d = 2.

5 1964 E L E M E N T A R Y P R O B L E M S A N D S O L U T I O N S 235 Solutio by B. Litvack, Uiversity of Michiga, A Arbor, Michiga; Jch H. Halto, Uiversity of Colorado, Boulder, Colorado; ad the proposer. S u p p o s e t h a t (1) a u + bv cw = 0 for = 1 a d = 2. M u l t i p l y i g the f i r s t c a s e by h a d the s e c o d by g, w e s e e t h a t a h u, + bhv, chw = 0, A d d i g, w e o b t a i a g u 2 + b g v c g w 2 = 0. au~ + bv« cw» s i c e u, v,..., w a i l s a t i s fj y y + 2 = ^ y h y. R e p e a t i g t h e p r o c e s s ( o r, m o r e f o r m a l l y, u s i g m a t h e m a t i c a l i d u c - tio) w e v e r i f y t h a t (1) h o l d s for a l l if it h o l d s for = 1, 2. A l s o s o l v e d J A R D E N P R O D U C T S Proposed by G.L. Alexaderso, Califoria Uiversity of Sata Clara, Sata Clara, L e t u a d v be a y two s e q u e c e s s a t i s f y i g t h e s e c o d - o r d e r r e c u r r e c e f o r m u l a < x > y + 2 = *y +1 + h v w h e r e g a d h a r e c o s t a t s. Show t h a t t h e s e q u e c e of p r o d u c t s w = u v s a t i s f i e s a t h i r d - o r d e r r e c u r r e c e f o r m u l a (2 > y + 3 = a y b y + l + C y a d fid a, b, a d c a s f u c t i o s of g a d h.

6 236 ELEMENTARY PROBLEMS AND SOLUTIONS October Solutio by the proposer. Let r ad s be the roots of the auxiliary polyomial x -gx-h of (1). We assume r j> s; the case r = s has the same result. Now,., %,. 2. u = C l l r + C 1 2 S ' V c 2 1 r + C 2 2 S ' a d s o V c l ( r ) + c 2 ( r s ) + c 3 ( s ) ' Hece the auxiliary polyomial of (2) is x -ax -6x-c=(x-r )(x-rs)(x-s ) = _x -(r +s )x+(rs) J (x-rs) = [x 2 -(g 2 +2h)x+h 2 ] (x+h)^x 3 -(g 2 +h)x 2 -(g 2 +h)hx+h Now a=g +h, b=(g +h)h, ad c=-h Also solved by Joh H. Halto, Uiversity of Colorado, Boulder, Colorado; ad Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas. This problem is a special case of formulas of D. Jarde, Recurrig Sequeces, Riveo Lematematika, Jerusalem (Israel), 1958, p. 43. This problem is a special case of formulas of D. Jarde, Recurrig Sequeces, Riveo Lematematika, Jerusalem (Israel), 1958, p. 43. AN ALTERNATING BINOMIAL TRANSFORM B Proposed by J. L. Brow, Jr., Pesylvaia State Uiversity, Uiversity Park, * Pesylvaia Prove that k=l for r a odd positive iteger ad geeralize. Solutio by H. H. Fers, Uiversity of Victoria, Victoria, British Columbia, Caada We have the Biet formula F q - P A where v?, a 1 - V5 a = - ad p =

7 1964 ELEMENTARY PROBLEMS AND SOLUTIONS 237 Thus 1 - a = or 1 - /3 = a. Now r - 1 k=l. - ^ ( W-o&rt <-"'-' G - ' D ^ 1 - ) \ - G > + «2 -G)«3 *-- + <- 1» r ~'(,! 1 >,r " 1 Let r be a odd positive iteger. The r-1. 2 ( - i ) k F k = ^ { f ( 1 - a ) r a r ] + C l - ( 1-0 ) r - ^ r ] / k=l 5 = I - [ J 3 r a r + l - a r - 3 r ] = 0. Let r be a eve positive iteger. The r-1 2 H) k fflf k - I{[(i - ) r - i - «r J + Ci. - a -*)* +0']} = I [ 0 r a r + i - a r + $. r ] -2 ( ^ ) -2F

8 238 ELEMENTARY PROBLEMS AND SOLUTIONS October For Lucas umbers it ca be show by aalogous methods that r-1 Z m*" (JK { k=l if r is eve 2-2L if r is odd. ^ r Also solved by Joh H. Halto, Uiversity of Colorado, Boulder, Colorado; Douglas Lid, Falls Church, Virgiia; Charles R. Wall, Texas Christi Uiversity, Ft. Worth, Texas; ad the proposer. THE P E L L S E Q U E N C E B Proposed by Roseaa Torretto, Uiversity of Sata Clara, Sata Clara, Califoria The sequece 1, 2, 5, 12, 29, 70,... is defied by c, = 1, c 0 = 2, ad c. = 2c,,. + c for all 2 1. Prove that c_ is a m itegral multiple of 29 for all positive itegers m. Solutio by Douglas Lid, Falls Church, Virgiia Sice c_ = 29, the solutio follows at oce from the m o r e gee r a l fact that for the above defied sequece, (1) x c c. ' m ' m We shall prove this m o r e geeral a s s e r t i o followig N. N. Vorobyov (The Fiboacci Numbers, Heath, 1963). We eed first establish that ( 2 ) C +k = C - l C k + C C k+l ' Proof is by iductio o k. The c a s e s k = 1, k = 2 a r e easily show t r u e. We the a s s u m e (2) true for k ad k + 1. Hece { 3 ) c +k = C - l C k + C C k+l * ( 4 ) C +k+l = C -l C k+l + C C k - 2 ' Multiplyig (4) by two ad addig to (3), we obtai C +k+2 = C -l C k+2 + C k+3"' completig the iductio step ad provig (2).

9 1964 ELEMENTARY PROBLEMS AND SOLUTIONS 239 We ow prove the geeral a s s e r t i o (1) by iductio usig (2). (1) is obviously true for = l. Nowassume c is divisible by c, 3 m m 9 2l 1 ad cosider c,,, v. By (2), (+l)m J \ t> (+l)m "" m-1 m m m+1 The first t e r m o the right is divisible by c hypothesis so is the last t e r m., ad by the iductio Applyig the fudametal theorem of a r i t h m e t i c, so also m u s t be c, I1X. This completes the iductio (+l)m ^ step ad the proof of (1). Also solved by B. Litvack, Uiversity of Michiga, A Arbor, Michiga; Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas; Joh H. Halto, Uiversity of Colorado, Boulder, Colorado; Dermott A. Breault, Sylvaia A.R.L., Waltham, Mass.; J.A.H. Huter, Toroto, Otario, Caada; H.H. Fers, Uiversity of Victoria, Victoria, British Columbia, Caada; J.L. Brow, Jr., Pesylvaia State Uiversity, State College, Pesylvaia; ad the proposer. HARMONIC DIVISION B Proposed by Brother U. Alfred, St. Mary's College, Califoria Give a lie with a poit of origi O ad four positive positios A, B, C, ad D with r e s p e c t to O. If the lie segmets OA, OB, OC, ad OD correspod respectively to four cosecutive Fiboacci umb e r s F, F,,, F, F l o, determie for which set(s) of Fiboacci u m b e r s the poits A, B, C, ad D a r e i simple harmoic ratio, i. e., AB AD, BC DC ' Solutio by Joh H. Halto, Uiversity of Colorado, Boulder, Colorado O, A, B, C, D a r e five cosecutive poits o a lie, with OA=F, OB=F, OC=F,,, OD=F ^. Thus AB=F X l - F =F., BC=F, F = F, AD= F ^. - F = (F,-, + F,, ) - (F x, * +2 +1' +2 - F +1',, ) = 2F +1,,, DC=F 0 - F ^ = - F ^. Thus AD/DC = -2, ad AB/BC =F, / F ' -L If B a d D divide A ad C harmoically, (AB/BC)(AD/DC) = - 1. That i s, F T / F = ^r. This occurs p r e c i s e l y oce, for positive, whe - 1 ' 2 * =3, ad ever for egative. The oly set of poits is therefore that i which OA=F 3 =2, OB=F 4 =3, OC=F 5 =5, OD=F 6 =8.

10 240 ELEMENTARY PROBLEMS AND SOLUTIONS October Editorial ote. Let R =F, / F. It is well kow ad easily - 1 ' ' proved that R 2 > R 4 > R, >... R > R > R 3. This shows that the for which R is uique. Also solved by Charles R. Wall, Texas Christia Uiversity, ad the proposer. Ft. Worth, Texas XXXXXXXXXXXXXXX Cotiued from page 184. Moreover, these a r e the dimesios of the cuboid of uit volume, for ip x 1 x<p = 1 _ 1, 1. a r e : 2. A -1 D H S? ' \ ' 1 1 I T Fig. 2 '/ F Certai other p r o p e r t i e s of the Golde Cuboid m a y be oted. It is clear from Fig. 2 that the ratios of the a r e a s of the faces AE:AC:CE = «p:l:«p"l. The total surface a r e a of the cuboid is 3( <p fp" 1 ) = 6*P 3. Four of the six faces of the cuboid a r e Gold Rectagles, e. g., CE (Fig. 3) 4. Each of the four diagoals of the cuboid is iclied to the base at a agle of The ratio of the a r e a of the sphere c i r c u m s c r i b i g the cuboid to that of the cuboid is 27r-.3<P. Oe further poit is of iterest. 6. It is well kow that, if a square CK is cut off from the Golde Rectagle CE (Fig. 3), the sides of the remaiig rectagle LE a r e also i the ratio <p:l. Ad of course the dissectio m a y be repeated util the rectagle size approaches that of a poit, which is the i t e r - sectio of BF ad KE. It is ot so well kow that, if two cuboids of square c r o s s s e c - tio (<?- *" x <P~*) a r e cut from the Golde Cuboid (broke lies, Fig. 2), the edge legths of the remaiig cuboid a r e i the same ratio as those of the origial cuboid, viz., 1: V" 1 : <p~ 2 = <p:l: <P~^, so that this also is a Golde Cuboid, ^p-3 times the size of the origial. The repetitio of the decapitatio p r o c e s s will lead to a idefiitely small Golde Cuboid located about a fixed poit. of this poit is left as a e x e r c i s e to # the r e a d e r. XXXXXXXXXXXXXXX The locatio